在平面直角坐标系xoy中,已知圆x^2+y^2=1与x轴正半轴的交点为F,AB为该圆一条弦,直线AB的方程为X=M,记以
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  • (1)圆心C(m,0),(-1<m<1),

    则⊙C的半径为:r=

    从而⊙C的方程为(x-m)2+y2=1-m2,

    椭圆D的标准方程为:

    (2)当b=1时,椭圆D的方程为

    x221+y2=1,

    设椭圆D上任意一点S(x1,y1),

    x122+y12=1,y12=1-

    ∵SC2=(x1-m) 2+y12

    =(x1-m) 2+1-

    = (x1-2m)2+1-m2

    ≥1-m2=r2,

    所以SC≥r.

    从而椭圆D上的任意一点都不存在⊙C的内部.

    (3)

    ​OL=b2+1为定值.

    证明:设点P(x1,y1),Q(x2,y2),

    则由题意,得N(x1,-y1),x1≠x2,y1≠±y2,

    从而直线PQ的方程为(y2-y1)x-(x2-x1)y+x2y1-x1y2=0,

    令y=0,得xM=

    x1y2-x2y1 y2-y1,

    ∵直线QN的方程为(y2+y1)x-(x2-x1)y-x1y2-x2y1=0,

    令y=0,得xL=

    x2y1+x1y2y2+y1.

    ∵点P,Q在椭圆D上,

    x12b2+1+

    y12b2=1,

    x22b2+1+

    y22b2=1,

    ∴x12=b2+1-

    b2+1b2y12,x22=b2+1-

    b2+1b2y22,

    ∴xM•xL=

    (b2+1-

    b2+1b2y12)y22-(b2+1-

    b2+1b2y22 )y12y22-y12

    =

    (b2+1)(y22-y12)y22-y12=b2+1.

    ​OM•

    ​OL=xM•xL=b2+1为定值.